Question: Simplify and expand the following expression: $ \dfrac{5z - 6}{z - 4}-\dfrac{5z - 6}{z - 1} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(z - 4)(z - 1)$ Multiply the first term by $\dfrac{z - 1}{z - 1}$ $ \begin{align*} \dfrac{5z - 6}{z - 4} \times \dfrac{z - 1}{z - 1} & = \dfrac{(5z - 6)(z - 1)}{(z - 4)(z - 1)} \\ & = \dfrac{5z^2 - 11z + 6}{(z - 4)(z - 1)}\end{align*} $ Multiply the second term by $\dfrac{z - 4}{z - 4}$ $ \begin{align*} \dfrac{5z - 6}{z - 1} \times \dfrac{z - 4}{z - 4} & = \dfrac{(5z - 6)(z - 4)}{(z - 1)(z - 4)} \\ & = \dfrac{5z^2 - 26z + 24}{(z - 1)(z - 4)}\end{align*} $ Now we have: $ = \dfrac{5z^2 - 11z + 6}{(z - 4)(z - 1)} - \dfrac{5z^2 - 26z + 24}{(z - 1)(z - 4)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5z^2 - 11z + 6 - (5z^2 - 26z + 24)}{(z - 4)(z - 1)} $ $ = \dfrac{5z^2 - 11z + 6 - 5z^2 + 26z - 24}{(z - 4)(z - 1)} $ $ = \dfrac{15z - 18}{(z - 4)(z - 1)}$ Expand the denominator: $ = \dfrac{15z - 18}{z^2 - 5z + 4}$